1. 原題： https://www.patest.cn/contests/pat-a-PRactise/1079

2. 思路：

3. 源碼：

#include<iostream>#include<vector>using namespace std;int N;//結(jié)點(diǎn)數(shù)double price, r, sum = 0;//分別為價(jià)格， 比率， 總價(jià)vector<int> pro;//存儲(chǔ)葉子結(jié)點(diǎn)的產(chǎn)品數(shù)量vector< vector<int> > G;//圖的鄰接表表示void dfs(int sn, double price);//dfs遞歸，第一個(gè)參數(shù)是根結(jié)點(diǎn)int main(void){	//freopen("in.txt", "r", stdin);	scanf("%d %lf %lf", &N, &price, &r);	pro.resize(N, 0);	G.resize(N);	for (int i = 0; i < N; i++)//讀入數(shù)據(jù)	{		int cnt;		scanf("%d", &cnt);		if (cnt == 0)			scanf("%d", &pro[i]);//存儲(chǔ)該結(jié)點(diǎn)的產(chǎn)品數(shù)量		else		{			for (int j = 0; j < cnt; j++)			{				int id;				scanf("%d", &id);				G[i].push_back(id);			}		}	}	dfs(0, price);//遞歸	printf("%.1lf/n", sum);	return 0;}void dfs(int sn, double price){	if (G[sn].empty())//表示葉子結(jié)點(diǎn)	{		double tem = pro[sn] * price;//累計(jì)價(jià)錢		sum += tem;		return;	}	price *= (1 + r / 100.0);	for (int i = 0; i < G[sn].size(); i++)//遞歸下一層		dfs(G[sn][i], price);	return;}